Language note: This is not a "sum convergence" problem. We say a sum converges of the sequence $a_0$, $a_0+a_1$, $a_0+a_1+a_2$,... converges. This problem is not a problem of that type. Rather this is a question about the limit of a sequence of different (related) sums.
–
Thomas AndrewsSep 29 '12 at 21:24

Hint: you can rewrite your sum as
$$
\lim_{n\to\infty} \frac{1-0}{n}\sum_{i=1}^n \frac{1}{1+\frac{i}{n}},
$$
Now do you know the definition of Riemann's integral?

Added: Somehow the english Wiki page doesn't not seem to show this as explicitly as the french one does, but you can have a look at this page, the first section shows what you have, with $f$ replaced by $\frac{1}{1+x}$

Is Riemann's integral that theorem they teach you in calc 1 when you take the number of rectangles under a curve to be infinite and compute their areas? If so, it has been a long long time! hahah
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CodeKingPlusPlusSep 29 '12 at 21:09

It is somewhat the way Riemann's integral is constructed in the first calculus course, via approximatio of areas under curves by rectangle.
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Jean-SébastienSep 29 '12 at 21:12

Could you show some derivation of how you get $\int\dfrac{dx}{1+x}$
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CodeKingPlusPlusSep 29 '12 at 21:34

Riemann sums, as simple as that. Since the function $\,f(x)=\frac{1}{1+x}\,$ is continuous in $\,[0,1]\,$ its Riemann integral exists there and we can thus choose any subdivision of $\,[0,1]\,$ we want to form the Riemann sums. We choose $\,\{x_0=0\,,\,x_1=1/n\,,\,x_2=2/n\,,...,\,x_n=n/n=1\}\,$ and we form the Riemann sum $$\sum_{i=1}^nf(x_i)(x_i-x_{i-1})=\sum_{i=1}^n\frac{1}{1+\frac{i}{n}}\cdot\frac{‌​1}{n}$$
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DonAntonioSep 29 '12 at 21:41